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Del congelador se saca 60 g de hielo a – 10 ºC. Determina la cantidad de calor necesario que debe absorber el hielo para obtener agua líquida a temperatura ambiente. (T ambiente = 20 \text{ ºC}, Patm = 1 \text{ atm})

","status","pending_classification","hasRelatedVideo","relatedVideos",[109],{"_110":111,"_129":130,"_51":145,"_163":164,"_181":146},"content",{"_43":112,"_57":113,"_114":-5,"_115":116,"_117":118,"_119":-5,"_45":120,"_123":124,"_125":124,"_126":-5,"_127":128},"162087be-594f-460e-bbdd-a339ac2f10ad","Calor Latente - Parte 1","description","content_type_id","1051eee7-c62e-4b2c-9900-77a29fcec080","resource_id","f2b62fc2-7079-40e7-a5b5-c4a9769b205d","user_id",{"_121":122},"external_id","3659","created_at","2025-08-20T12:27:41.608438+00:00","updated_at","content_group_id","slug","calor-latente-parte-1","prometheusData",{"_131":132,"_138":139,"_140":141,"_142":143,"_28":144},"provider",{"_43":133,"_134":135,"_136":137},"06f120be-2194-4548-8336-7ba44ffbae14","key","grupo_q_proteus","kind","proteus","reference","bf8aacba-3de9-4b10-961c-47ada80dfe8c","state","completed","thumbnail_url","https://proteus.grupoq.io/videos/bf8aacba-3de9-4b10-961c-47ada80dfe8c/thumbnailsthumb.0000000.jpg","https://proteus.grupoq.io/videos/bf8aacba-3de9-4b10-961c-47ada80dfe8c/levels.m3u8",[146],{"_43":147,"_57":148,"_59":149,"_150":151,"_152":153,"_123":154,"_125":154,"_45":155,"_127":149,"_161":162},"96193fa0-9495-45b2-81cc-b547604354af","Física Térmica - Termologia","fisica-termica-termologia","tag_group_id","c1ce2cf2-b217-4cc7-8182-81c56bd3522d","parent_id","c8ab2a95-fd84-4d6f-ade0-835d34ba00f9","2025-04-29T03:44:27.560682+00:00",{"_156":157,"_158":64,"_159":160},"childrenCount",10,"content_count","question_count",8575,"children",[],"video",{"_43":118,"_13":113,"_114":-5,"_165":166,"_167":168,"_28":-5,"_169":-5,"_131":168,"_170":171,"_142":-5,"_172":-5,"_173":-5,"_174":-5,"_175":-5,"_45":176,"_177":64,"_178":-5,"_104":179,"_123":180,"_125":180},"duration",17,"source_type","prometheus","iframe_url","provider_video_id","748a4adb-ad8f-4c5f-88a9-bc522d3b4246","format","quality","file_path","file_size",{"_121":122},"views_count","last_played_at","draft","2025-08-19T11:56:22.11908+00:00","tag","similarQuestions",[184,214,232],{"_40":185,"_42":41,"_43":186,"_77":187,"_101":202,"_102":203,"_51":204},"112831da-be9c-458e-94cb-e8292d0dba12","3c0b9441-b878-4896-8bac-fa4ca61e3f13",[188,191,194,197,200],{"_43":189,"_81":82,"_83":190},"265bd71a-6749-41f5-b291-67efd646ca64","y'' + 2y' - 3y = 0",{"_43":192,"_81":87,"_83":193},"6d3c3a1d-8e80-4ff3-b97a-d5b5bca91940","6y'' - y' - y = 0",{"_43":195,"_81":91,"_83":196},"4168ac67-64fd-46ed-a69c-a01dbfcde7ea","y'' - 2y' + y = 0",{"_43":198,"_81":95,"_83":199},"cc7bc8df-6995-4264-ac5a-32ea9b4ed7db","9y'' + 6y' + y = 0",{"_43":185,"_81":99,"_83":201},"y'' - 2y' + 6y = 0","Somente em y''−2y'+6y=0 o polinômio característico r²−2r+6=0 tem Δ<0, gerando raízes complexas conjugadas (1±i√5) e, portanto, solução em termos de exponencial modulando seno e cosseno. Nos demais itens (a–d) o discriminante é ≥0, produzindo raízes reais (distintas ou repetidas).","

Resolva as seguintes EDOs determinando a solução geral e, quando houver condições iniciais do PVI, a solução particular.

(a) y'' + 2y' - 3y = 0

(b) 6y'' - y' - y = 0

(d) 9y'' + 6y' + y = 0

(e) y'' - 2y' + 6y = 0

(f) 4y'' + 9y = 0

(g) 6y'' - 5y' + y = 0; \\, y(0) = 4; \\, y'(0) = 0

(h) y'' + 4y' + 4y = 0; \\, y(-1) = 2; \\, y'(-1) = 1

(i) y'' + y = 0; \\, y\bigg(\\frac{\\pi}{3}\\bigg) = 2; \\, y'\\bigg(\\frac{\\pi}{3}\\bigg) = -4

(j) y'' - 2y' - 3y = 3e^{2t}

(k) y'' + 9y = t^2 e^{3t} + 6

(l) 2y'' + 3y' + y = t^2 + 3 \\sin(t)

(m) y'' + y' - 2y = 2t; \\, y(0) = 0; \\, y'(0) = 1

(n) y'' - 2y' + y = te^t + 4; \\, y(0) = 1; \\, y'(0) = 1

(o) y'' + 4y = 3 \\sin(2t); \\, y(0) = 2; \\, y'(0) = -1

",[205],{"_43":206,"_57":207,"_59":208,"_150":151,"_152":209,"_123":210,"_125":210,"_45":211,"_127":208},"6516bbbd-f2a3-4710-8382-702240b17836","Mecânica na Engenharia de Petróleo","mecanica-na-engenharia-de-petroleo","b787e438-0ab3-454c-ab43-5d3d14c2720d","2025-04-29T03:44:35.328997+00:00",{"_156":212,"_158":64,"_159":213},4,12302,{"_40":215,"_42":41,"_43":216,"_77":217,"_101":229,"_102":230,"_51":231},"6831a9b9-c9a4-4a76-bd7a-2ce98bfea875","93a32dd4-5fc6-4ccb-a76f-4724bf113742",[218,221,224,226],{"_43":219,"_81":82,"_83":220},"13e3f476-a0a5-421f-b47e-0969e8e9a7e0","

Verdadeiro, verdadeiro, falso

",{"_43":222,"_81":87,"_83":223},"c9a20855-d84b-4860-8b35-9784818282cf","

Falso, verdadeiro, verdadeiro

",{"_43":215,"_81":91,"_83":225},"

Verdadeiro, falso, verdadeiro

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Falso, falso, falso

","85,5 % decomposto → 14,5 % restante.\nN/N0 = 0,145 ⇒ t = ln(1/0,145)/(ln2/5730) ≈ 1,6×10⁴ anos.\nItens prováveis: (1) resta 14,5 % do C-14 (V); (2) idade ≈ 1,2×10⁴ anos ou 3 meias-vidas (F); (3) t½ do C-14 = 5730 anos (V). Sequência: V-F-V, correspondendo à alternativa c.","

Qual é a idade da madeira na qual 85,5\\% do C-14 se decompos?

",[205],{"_40":233,"_42":41,"_43":234,"_77":235,"_101":250,"_102":251,"_51":252},"3ce109b1-6cdb-4b4d-8816-a6af4d74bed0","c478a46c-46a8-4942-ae14-2e79def3dda2",[236,239,242,244,247],{"_43":237,"_81":82,"_83":238},"a5a16a42-3e58-4c76-80f2-c291aa2d8e49","Sim",{"_43":240,"_81":87,"_83":241},"5e6d0254-696c-4da0-b572-82e3a19f433e","Não",{"_43":233,"_81":91,"_83":243},"Indeterminado",{"_43":245,"_81":95,"_83":246},"35403f68-a880-429d-97f4-db507c58de38","Todos",{"_43":248,"_81":99,"_83":249},"2e06d948-d37e-4350-afc7-ced04941bcaf","Nenhum","Os polinômios e números fornecidos estão corrompidos (símbolos “????”), impossibilitando aplicar Ruffini e concluir se são raízes. Sem dados completos, o problema é indeterminado.","Emplea la regla de Ruffini para dictaminar si los siguientes números son o no raíces de los polinomios citados:
a) ????3 de ????4????5
b) ????2 de ????2????????2
c) ????1 de 2????????1
d) ????1 de 2????2????",[205],[205,254,261,271,278,285,286,294,300,305,311,318],{"_43":255,"_57":58,"_59":60,"_150":151,"_152":256,"_123":154,"_125":154,"_45":257,"_127":260},"6b6faf95-0e3a-4f58-abc8-e1444c48041d","92339f4f-ed20-4f4c-ad85-0c3de7e04ae0",{"_156":258,"_158":64,"_159":259},2,11846,"termodinamica-3",{"_43":262,"_57":263,"_59":264,"_150":151,"_152":265,"_123":266,"_125":266,"_45":267,"_127":270},"861ad829-750d-4a44-9569-2f18b430b4df","Engenharia Mecatrônica","engenharia-mecatronica","eb9315c5-bd04-4116-aa46-02c4fde320c0","2025-04-29T02:19:12.828432+00:00",{"_156":268,"_158":64,"_159":269},5,30614,"engenharia-mecatronica-1",{"_43":272,"_57":273,"_59":274,"_150":151,"_152":262,"_123":210,"_125":210,"_45":275,"_127":277},"88fe96b3-2a17-4198-90d5-ddda3d6f5d45","Mecânica","mecanica",{"_156":268,"_158":64,"_159":276},25741,"mecanica-3",{"_43":256,"_57":279,"_59":280,"_150":151,"_152":265,"_123":266,"_125":266,"_45":281,"_127":284},"Engenharia Mecânica","engenharia-mecanica",{"_156":282,"_158":64,"_159":283},11,36318,"engenharia-mecanica-2",{"_43":147,"_57":148,"_59":149,"_150":151,"_152":153,"_123":154,"_125":154,"_45":155,"_127":149},{"_43":287,"_57":288,"_59":289,"_150":151,"_152":-5,"_123":290,"_125":290,"_45":291,"_127":289},"9772e958-72c6-4499-ad47-877cf94d670d","Exatas","exatas","2025-04-29T01:56:45.284172+00:00",{"_156":292,"_158":64,"_159":293},18,135489,{"_43":295,"_57":58,"_59":60,"_150":151,"_152":206,"_123":296,"_125":296,"_45":297,"_127":299},"b21374de-aae5-4a4a-9bf3-0420e731797e","2025-04-29T04:00:14.680867+00:00",{"_156":64,"_158":64,"_159":298},8597,"termodinamica-5",{"_43":209,"_57":301,"_59":302,"_150":151,"_152":265,"_123":266,"_125":266,"_45":303,"_127":304},"Engenharia de Petróleo","engenharia-de-petroleo",{"_156":268,"_158":64,"_159":213},"engenharia-de-petroleo-1",{"_43":153,"_57":306,"_59":307,"_150":151,"_152":287,"_123":266,"_125":266,"_45":308,"_127":310},"Física","fisica",{"_156":282,"_158":64,"_159":309},13777,"fisica-4",{"_43":265,"_57":312,"_59":313,"_150":151,"_152":-5,"_123":314,"_125":314,"_45":315,"_127":313},"Engenharia","engenharia","2025-04-29T01:55:38.635246+00:00",{"_156":316,"_158":64,"_159":317},29,351586,{"_43":319,"_57":58,"_59":60,"_150":151,"_152":272,"_123":320,"_125":320,"_45":321,"_127":322},"f5967cbe-d751-4b40-81f2-04eb5dead2a9","2025-04-29T04:00:13.215003+00:00",{"_156":64,"_158":64,"_159":298},"termodinamica-4","tagsTree",[325,329],{"_43":287,"_57":288,"_59":289,"_150":151,"_152":-5,"_123":290,"_125":290,"_45":291,"_127":289,"_161":326},[327],{"_43":153,"_57":306,"_59":307,"_150":151,"_152":287,"_123":266,"_125":266,"_45":308,"_127":310,"_161":328},[146],{"_43":265,"_57":312,"_59":313,"_150":151,"_152":-5,"_123":314,"_125":314,"_45":315,"_127":313,"_161":330},[331,337,341],{"_43":262,"_57":263,"_59":264,"_150":151,"_152":265,"_123":266,"_125":266,"_45":267,"_127":270,"_161":332},[333],{"_43":272,"_57":273,"_59":274,"_150":151,"_152":262,"_123":210,"_125":210,"_45":275,"_127":277,"_161":334},[335],{"_43":319,"_57":58,"_59":60,"_150":151,"_152":272,"_123":320,"_125":320,"_45":321,"_127":322,"_161":336},[],{"_43":256,"_57":279,"_59":280,"_150":151,"_152":265,"_123":266,"_125":266,"_45":281,"_127":284,"_161":338},[339],{"_43":255,"_57":58,"_59":60,"_150":151,"_152":256,"_123":154,"_125":154,"_45":257,"_127":260,"_161":340},[],{"_43":209,"_57":301,"_59":302,"_150":151,"_152":265,"_123":266,"_125":266,"_45":303,"_127":304,"_161":342},[343],{"_43":206,"_57":207,"_59":208,"_150":151,"_152":209,"_123":210,"_125":210,"_45":211,"_127":208,"_161":344},[345],{"_43":295,"_57":58,"_59":60,"_150":151,"_152":206,"_123":296,"_125":296,"_45":297,"_127":299,"_161":346},[],"videoDislikes",{},"videoLikes",{},"actionData","errors"]

Comentários